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In Zorich’s mathematics analysis, a theorem states that “Suppose $M\subset \mathbb{R}^{n}$, $f$ : $M\rightarrow\mathbb{R}$ is continuous, then the set$\{(x_1, x_2,…, x_n,y): y=f(x_1, x_2,…, x_n), (x_1, x_2,…, x_n)\in M\}$ is Lebesuge measure zero in $\mathbb{R}^{n+1}$.” How to prove this theorem? First I thought that there is a function $g$ : $\mathbb{R}^{n}\rightarrow\mathbb{R}$ such that $g$ is continuous, and its restriction on $M$ is $f$. Then the diagram of $f$ is a subset of the diagram of $g$, hence measure zero. However, later I found that this proof may not work because such $g$ can not always be constructed. For example simply consider $f(x)=\tan x$. So how can we prove this statement?(Since I have only been learning basic caculus, without much knowledge in measure theory, I hope that the proof only requires basic caculus) Remark:A set $S$ is Lebesuge measure zero, if $\forall \epsilon>0$, $\exists$ at most countable close intervals, with their union containing $S$, and sum of their volumes less than $\epsilon$.

Asigan
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  • This is an immediate consequence of Fubini's Theorem since every singleton set has measure $0$. – geetha290krm Jul 07 '22 at 09:53
  • @geetha290krm Well, do you have a proof directly using the definition of measure zero? I have not learned much measure theory. – Asigan Jul 07 '22 at 10:31
  • See Graph of a continuous function has measure zero. – Dave L. Renfro Jul 07 '22 at 10:57
  • @Dave L. Renfro Yes, I have read this post. But this post only solves when $M=\mathbb{R}^n$. And a key step in one proof not using the measure theory uses the fact that $\mathbb{R}^{n}$ is countable union of compact sets, then applying uniform continuity. But is $M$ countable union of compact sets for general $M$? – Asigan Jul 07 '22 at 13:06
  • $M$ does not have to be $\sigma$-compact, but unless I'm overlooking something, simply use the fact that $M$ is contained in a $\sigma$-compact set, and obtain appropriate coverings associated with this (possibly) larger set, which will then be appropriate coverings associated with $M.$ – Dave L. Renfro Jul 07 '22 at 13:14
  • @DaveL.Renfro Do you mean that I should show the diagram of the restriction of $f$ on $M\cap K$, where $K$ compact, is measure zero? But how to show that? – Asigan Jul 08 '22 at 08:35
  • It seems having an arbitrary subset $K$ (allows the function to be unbounded in many places) caused similar problems in the essentially identical MSE question Is the graph of a continous function $f:E\to\mathbb{R}$($E\subset\mathbb R^{n-1}$) a set of n-dimentional measure zero? (which I found by this google search). Maybe show measure zero for the intersection of the graph with every rectangle in ${\mathbb R}^{n+1}$? – Dave L. Renfro Jul 08 '22 at 10:44
  • Of course, there are uncountably many rectangles in ${\mathbb R}^{n+1},$ but you can cover ${\mathbb R}^{n+1}$ with at most countably many rectangles, say $R_1,$ $R_2,$ $R_3,;\ldots$ Then, assuming you can show the intersection of the graph with each of these rectangles has measure zero, given $\epsilon > 0,$ there exists a countable cover of the intersection with $R_1$ whose volumes sum to less than $\frac{1}{2}\epsilon,$ and there exists a countable cover of the intersection with $R_2$ whose volumes sum to less than $\frac{1}{4}\epsilon,; \ldots$ – Dave L. Renfro Jul 08 '22 at 10:51
  • @DaveL.Renfro Thanks, this is indeed helpful. – Asigan Jul 11 '22 at 08:34

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